Interview guides · AAA Canada

AAA Canada Software Engineer Interview Questions & Guide 2026

AAA Canada software engineer prep: algorithm drills where every complexity claim has to name the assumption under it.

The 60-second brief

As a Software Engineer (often categorized internally as a Methods Engineer) at AAA Canada, you are tasked with bridging the gap between technical strategy and operational excellence. This role is critical to the organization’s ability to maintain high standards of service and efficiency. You will be responsible for analyzing, designing, and refining the processes and software solutions that support AAA Canada's core business operations.

Ask how many rounds there are and what each one is before you plan your weeks, because the composition is what your hours should follow. A loop with three coding rounds and one short design conversation deserves a different split from the reverse, and whoever schedules it will usually say plainly which it is.

Focus areas

Bound every outbound call with a timeout
Paginate large result sets with keyset cursors
Choose indexes from the query's access path

Overview

The work you perform directly affects AAA Canada's technical infrastructure and the reliability of its service delivery. You will operate in a fast-paced environment where your ability to translate complex requirements into actionable technical plans is paramount. Whether you are optimizing existing workflows or contributing to new project initiatives, your contribution ensures that AAA Canada remains at the forefront of its industry.

Where to focus your preparation

Treat the sections below as preparation areas and confirm the format with your recruiter.

Preparation focus

No round sequence has been reported for this company, so work the categories below and confirm the format with your recruiter.

What to demonstrate

Breadth across SQL, experimentation and product reasoning
Ability to state assumptions before choosing a method

How to prepare

Drill the practice exercises below and time yourself
Prepare three quantified stories about decisions you drove

Where candidates lose points

Paginating with LIMIT/OFFSET over a set that changes while the client is reading it

OFFSET n makes the database produce and discard n rows before returning anything, so the cost of a page grows with its depth rather than with its size and page 500 costs five hundred pages of work. The correctness problem is worse than the cost: if a row is inserted or reordered between two page fetches, rows shift across the offset boundary and are either skipped entirely or returned twice, and neither outcome leaves any trace in the response for the client to detect. Keyset pagination - WHERE (sort_key, id) < ($last_sort_key, $last_id) ORDER BY sort_key DESC, id DESC LIMIT n, backed by an index in exactly that order - reads only the rows it returns and is stable against concurrent inserts. It requires the tie-break column: a timestamp is not unique, and duplicate sort keys straddling a page boundary reintroduce the skip it was adopted to remove.

Shipping a migration and the code that depends on it as a single change

During any rolling deploy, and for as long as a rollback remains possible, old and new code execute against the same schema at the same time. A migration that drops or renames a column breaks every instance that has not restarted yet, and code that requires a column the migration has not applied breaks every instance that restarted early. The discipline is expand then contract: add the new column nullable, write both shapes, backfill in batches, move reads across once the backfill is verified, and only then stop writing the old shape and drop it - four deploys, usually spread over days. It feels disproportionate until the first rollback, at which point it is the only reason the previous version still runs.

Comparing floating-point values for equality, or holding money in them

Binary floating point cannot represent 0.1 exactly, so repeated addition drifts and an equality check fails on values that are mathematically equal. Store currency as integer minor units or a decimal type, and compare floats against a tolerance you chose for a stated reason.

Check-then-act on shared state

Read, decide, write is not safe under concurrency unless the decision and the write are one atomic step: a unique constraint with conflict handling, a compare-and-set, or a row lock held for the whole transaction. Two requests can both pass the existence check before either inserts, which shows up as duplicate rows under load and never in a single-threaded test.

Questions & practice

Diff a projection against the primary without per-row point reads

Hard · reconciliation · range hashing · throttling

The listing projection has drifted and some rows show a stale version. The primary holds 40,000,000 resource rows across 12,000 tenants while serving 1,200 writes and 14,000 reads per second. The obvious repair, reading each resource row and comparing its version against the projection, is correct and would eventually finish. Explain precisely why it is unacceptable here, then give a diff that finds the differing rows, state its complexity, and make it safe to run against a live primary. Replication lag is usually under 100 ms and is not bounded.

Approach

Quantify the naive cost rather than calling it slow: 40,000,000 point reads at even 0.5 ms each is over five hours serialised, and the only lever is concurrency, which is exactly what you cannot spend. The primary's pool is sized for the write path, and 40,000,000 random reads evict the buffer cache that sustains the 85 percent cache hit rate, so the audit degrades the system it is auditing.
Replace random access with one ordered pass per side. Both sides can be read in (tenant_id, resource_id) order, which is a sequential scan on each and a merge join in O(n) time and O(1) memory. For a dense diff that is the whole answer, and it reads the primary once instead of 40,000,000 times.
For the expected sparse case, compare range hashes instead of rows: partition the key space, compute per range an order-independent aggregate over hash(resource_id, version), compare aggregates, and descend only into ranges that differ. With d differing rows and branching factor B, at most d ranges mismatch per level, so the drill-down examines O(d log_B(n/d)) ranges and reads full rows only in mismatching leaves.
Aggregate with a sum modulo 2^64 or a multiset hash, never XOR. XOR is order-independent but self-cancelling, so two rows wrong in the same way, or a row duplicated on one side, leave the range aggregate matching and the range is declared clean.
Pin the comparison to a point in time or it reports lag as drift: consider only rows whose updated_at is older than now minus a lag margin, and re-check each candidate mismatch individually before repairing. At 1,200 writes per second a diff without this reports thousands of false positives, and an unattended repairer would then overwrite live rows with stale values.
Make the run resumable and throttled: batch by range key, persist the last completed range, and watch a signal such as replica lag or primary CPU, pausing rather than pressing on. A reconciliation that cannot be stopped and resumed gets killed halfway and restarted from zero, which is how a repair becomes an incident.

Follow-up

The diff reports 900 stale rows. How do you decide between patching those rows and rebuilding the projection from resource_revision?
Same job, but the projection lives in a search index that cannot be scanned in key order. What changes?
How would you run this continuously at low cost instead of only as incident response?

Archive a resource graph without breaking live references or recursing

Medium · graph traversal · topological order · tenant isolation

Resources reference other resources within a tenant; for the largest tenant the reference table holds up to 2,000,000 nodes and 8,000,000 edges. Archiving a resource must archive everything reachable from it that nothing outside the set still references, refuse when a live external referrer exists, and terminate when references form cycles, which they legitimately do. Produce the archive order and the refusal list, targeting O(V+E). Say what stops the traversal crossing a tenant boundary, and why recursion is the wrong control structure at this size.

Approach

Load the subgraph with the tenant predicate on both endpoints of the edge, not only on the side you started from. Scoping the left table alone is the classic cross-tenant leak: one mis-entered edge then pulls another tenant's resources into the traversal and, worse, into the archive.
Traverse iteratively with an explicit stack. A 2,000,000-node graph can hold a chain deep enough to exhaust a native stack in the low tens of thousands of frames, and that failure is a process crash rather than an error you can return.
Treat cycles as data rather than corruption: compute strongly connected components with Tarjan in O(V+E) using its own explicit stack, then condense. The condensation is a DAG, so a topological order over it gives the archive order, and every member of a component archives in one transaction because no order within a cycle is valid.
Decide refusals with reverse edges. A candidate is archivable only if every in-edge originates inside the candidate set, so build the transpose or count in-degrees restricted to the visited set, and emit each blocked resource with the id of the external referrer, which is the only part of the answer an operator can act on.
Store the graph as CSR rather than a map of lists: an offsets array of V+1 8-byte entries plus E 8-byte targets is about 80 MB at this size, where boxed adjacency lists cost several times that and lose cache locality on every hop.
Run Kahn over the condensation for the order in O(V+E). If the emitted count is short of the component count the condensation step itself is wrong, since a condensation cannot contain a cycle, which makes the check free.

Worked solution (30 min)

Write the edge-loading query with the tenant predicate on both endpoints and state what it does with a cross-tenant edge.
Implement iterative Tarjan with an explicit stack and confirm on a three-node cycle that it emits one component of size three.
Build the transpose restricted to the visited set and mark every node with an in-edge from outside it as refused, carrying the referrer id.
Run Kahn over the condensation and verify the emitted order against the referrer-before-referenced rule.
Size the CSR arrays for 2,000,000 nodes and 8,000,000 edges and compare against a boxed adjacency map.

Expected result: An iterative O(V+E) traversal over a tenant-scoped CSR subgraph, SCC condensation so cycles archive atomically as one component, a transpose-based refusal list naming the external referrer for each blocked resource, and a Kahn topological order over the condensation, with recursion replaced by an explicit stack because of graph depth rather than style.

Verify it yourself

A two-node cycle with no external referrers archives as one unit, and adding a single external referrer moves both nodes to the refusal list.
No resource appears in the archive order before something that references it, except within its own component.
A counter on edge visits matches the loaded edge count exactly, confirming each edge is traversed once.

Follow-up

The graph is read in one query and the archive writes a minute later. What can change in between, and how do you make the write safe?
The candidate set is 400,000 resources. Is that one transaction, and if not, what does a half-finished archive look like to a reader?
An edge points at a resource in another tenant. Is that a refusal, an error, or an alert?

Track a rolling failure rate per destination for circuit decisions

Easy · sliding window · ring buffer · circuit breaker

The egress service delivers about 1,500 webhooks per second across roughly 40,000 destinations, each call bounded by a 10 second timeout. Maintain, per destination, the failure rate over the trailing 60 seconds so a caller can ask before dispatch whether the circuit should open. Attempts arrive as (destination_id, finished_at_ms, outcome). Requirement: amortised O(1) per attempt, with total memory bounded by the destination count rather than by traffic. Give the structure, its exact memory, and the rule that stops a destination with three attempts from opening a circuit.

Approach

Name the exact-deque version and then reject it as the default. Holding timestamps and advancing a tail pointer past anything older than now minus 60 seconds is a correct two-pointer window at amortised O(1) per attempt, but its memory tracks in-window traffic, so one destination in a retry storm holds hundreds of thousands of entries while thousands of quiet destinations hold none.
Use a ring of 60 one-second buckets per destination, each bucket a pair of counters for attempts and failures. On an attempt, advance the ring by the elapsed whole seconds, zeroing at most min(elapsed, 60) buckets, then increment the head. That is amortised O(1) with a fixed footprint per destination.
State the footprint: 60 buckets times two 4-byte counters is 480 bytes of payload per destination, so 40,000 destinations is roughly 20 to 25 MB with per-entry overhead, bounded by the catalogue rather than by the rate. The cost is granularity, since the oldest bucket ages out in whole seconds, which is far tighter than the decision needs.
Require a minimum sample before the circuit may open. A destination with three attempts and three failures reads as 100 percent and is not evidence; a floor of roughly 20 attempts in the window makes the ratio meaningful, and below that floor use a run of consecutive failures as the trigger instead.
Expire idle destinations, or memory grows with every destination ever seen rather than with the live set. Hold the rings in a bounded LRU keyed on destination_id and treat a miss as no history, which is the correct default for an endpoint that has been silent for a minute.
Keep the half-open probe out of the window arithmetic. After the circuit opens, one probe per interval decides whether to close it, and folding that single success into a window that still holds a 100 percent failure history would reopen the destination on one data point.

Follow-up

The fleet is 30 instances and each sees roughly a thirtieth of a destination's traffic. Where does the rate actually live, and what does a per-instance answer get wrong?
A destination answers in 9.5 seconds and succeeds. It is not failing but it is consuming your per-destination concurrency. What signal should open the circuit here?
How would you make the window survive a process restart, and is it worth the cost?

Replace offset paging on the resource feed with keyset

Easy · keyset pagination · composite index · sort order

resource holds resource_id, tenant_id, owner_user_id, title, body_ref, version, status ('draft','active','archived','deleted'), created_at, updated_at, deleted_at, with an index on (tenant_id, status, updated_at DESC, resource_id DESC). The listing endpoint returns active resources for one tenant, newest update first, 50 per page, today with LIMIT 50 OFFSET n. Tenants reach page 400 and rows are created while they read. Write the keyset query, define what the cursor carries and how it is encoded, and say which part of the index each predicate uses. Assume PostgreSQL 16.

Approach

Name the two failures separately. OFFSET 20000 makes the server produce and discard 20,000 rows, so page cost grows with depth rather than with page size. Independently, any write that changes how many rows sort above the offset moves the window between two fetches, and the direction decides which anomaly you get: an insert lands at the head of updated_at DESC and pushes already-returned rows down past the boundary, so they are returned a second time; a delete above the offset, or a row whose updated_at is bumped above the cursor, pulls rows up and one is never returned at all. Nothing in the response reveals either.
Write the seek: WHERE tenant_id = $1 AND status = 'active' AND (updated_at, resource_id) < ($2, $3) ORDER BY updated_at DESC, resource_id DESC LIMIT 50. The row-value comparison is one index range rather than a disjunction, and both columns are NOT NULL, which is what makes that comparison well defined.
Map each predicate onto the index: tenant_id and status are equality on the leading columns, (updated_at, resource_id) is the range, and the ORDER BY matches the index order so no Sort node appears and the scan stops after 50 rows. The DESC in the definition only matters for mixed directions — a plain ascending btree on the same columns is read backwards for this query.
Put both sort columns in the cursor and nothing the client can tamper with into another tenant: base64 of (updated_at, resource_id), validated server-side, with tenant_id taken from the principal.
State the residual honestly. Keyset is stable against concurrent inserts and deletes, but not against a row whose updated_at changes mid-scroll — that row moves in the ordering and can be seen twice. If the feed must be a snapshot, order by an immutable key or bound the page set with updated_at <= the cursor's start value.
Keep a total out of the page path. A tenant-wide COUNT(*) is the scan keyset just removed; fetch LIMIT 51 and return has_more instead.

Worked solution (25 min)

Seed one tenant with 500k active resources, 2% of them sharing an identical updated_at.
Time LIMIT 50 OFFSET 0 against OFFSET 20000 and record rows-read from EXPLAIN (ANALYZE, BUFFERS) for each.
Page the whole set with the keyset query while an insert-only writer adds 100 rows/second, collecting resource_ids, and repeat the run with OFFSET.
Repeat both runs under a second writer profile that also deletes 20 rows/second from pages already returned and bumps updated_at on 20 more, and diff each collected id set against the rows that existed for the whole run.
Remove resource_id from the cursor so the seek degrades to updated_at < $2, and re-run the tie-heavy section of the feed.

Expected result: Keyset reads about 50 index entries per page and its latency is flat from page 1 to page 10,000; under both writer profiles the collected ids contain no duplicate and omit no row that existed for the whole scroll. OFFSET at page 400 reads 20,050 rows to return 50. Under the insert-only writer it repeats rows and skips none — an insert can only push a row down past the boundary — and only the second profile, whose deletes and updated_at bumps move rows up past the cursor, also makes it skip. Without the tie-break, the seek is updated_at < $2, which drops every still-unread row sharing the boundary timestamp.

Verify it yourself

EXPLAIN shows an Index Scan with no Sort node — a Sort means the ORDER BY is not satisfied by the index.
Buffers read per page stays constant as page depth grows.
Ids collected across the scroll are unique, and complete for rows untouched during the run.

Follow-up

The client asks for 'jump to page 400'. What do you offer instead, and what does the honest version cost?
Sort order becomes user-selectable across four columns. How many indexes is that, and which would you refuse to add?
What does the cursor do when the row it points at has since been deleted?

Keep soft-deleted accounts from blocking re-registration

Easy · soft delete · unique constraints · engine differences

app_user holds user_id, tenant_id, email CITEXT, password_hash (NULL for SSO principals), email_verified_at, auth_version, status ('invited','active','suspended','deactivated'), created_at, updated_at, deleted_at. Two live accounts for one address inside a tenant must be impossible, but an address freed by a soft delete must be reusable, and the same tenant may delete and re-register it repeatedly. Write the uniqueness DDL for PostgreSQL 16, then the equivalent for MySQL 8 where partial indexes do not exist, and say what each permits once three deleted rows already hold that address.

Approach

Start from what is actually unique: not (tenant_id, email), but (tenant_id, email) among live rows. PostgreSQL says that directly — CREATE UNIQUE INDEX app_user_live_email ON app_user (tenant_id, email) WHERE deleted_at IS NULL. A full constraint over the same two columns burns the address permanently the first time someone deletes an account.
Keep case-insensitivity in the type or the index, never in the application: CITEXT as given, or UNIQUE (tenant_id, lower(email)) as an expression index where the extension is unavailable. A case-sensitive unique column is exactly how two accounts for one human appear.
For MySQL 8 the predicate has to move inside the key: add a discriminator column that is a constant 0 while the row is live and is set to user_id on delete, with UNIQUE (tenant_id, email, deleted_marker). Live rows share the constant and still collide; deleted rows differ from each other and stop colliding.
State the NULL variant and its dependency: leaving the marker NULL for deleted rows also works, because a unique index treats NULLs as distinct — true in MySQL, and true in PostgreSQL only under the default NULLS DISTINCT, which PostgreSQL 15 lets you reverse. Check the polarity against the three existing deleted rows: constant-on-live is what preserves the collision you want, and reversing it silently admits duplicate live accounts.
Say what a soft delete must do besides setting deleted_at: increment auth_version so existing tokens stop validating, leave resource.owner_user_id and resource_revision.actor_user_id intact, and accept that the address is retained — erasure is a different requirement answered by scrubbing the column, not by a DELETE that would break those references.

Follow-up

A deleted account re-registers with the same address the next day. Do the old resource rows follow the new user_id, and how does the API keep the two principals apart?
How do you honour an erasure request while resource_revision.actor_user_id still references this table?
What changes if a user may hold membership in two tenants?

Rebuild the search projection while it serves nine thousand queries

Medium · projections · eventual consistency · rebuild · watermark

The read-model service answers about 9k queries/second at a 120 ms p99 from a projection built off the event log, normally under 2 seconds behind. A mapping change forces a full rebuild from resource_revision, during which apply lag rises to minutes. Writes continue at 1.2k/second and events at 4k/second. Design the rebuild: how the new index is populated and cut over, how position is tracked per log partition, what the API returns alongside results so a client can tell a stale answer from a current one, and the criterion for cutting over.

Approach

Build into a second index and swap an alias rather than mutating the live one. The rebuild is then reversible by pointing the alias back, so a bad mapping costs a wasted rebuild instead of an outage. The price is peak storage for two full copies and double apply load during catch-up, and both numbers should be stated up front rather than discovered when disk fills.
Track position per log partition, not globally. Order is guaranteed only within an aggregate's partition, so progress is a vector, and the only number safe to publish is taken from the least advanced partition - the slowest one is what bounds completeness. Publishing the most advanced partition's position declares the projection current while another partition sits twenty minutes behind.
Make apply idempotent so the backfill and the live tail can overlap without a freeze. Each document records the aggregate_version it reflects, and any event at or below that version is discarded. This is why the event carries the full fact and not a delta: a delta cannot be discarded safely, and a consumer that calls back to read current state applies a state newer than the event it is processing, which is how a projection ends up with changes applied out of order.
Turn lag into a contract instead of a surprise. Return the watermark with every result set, and return the version produced by a write so the client can compare the two. A client that wrote version 7 and receives results at a watermark older than its own commit can show that its change is still landing, rather than rendering the previous value as current. Blocking the read until the projection catches up would convert a staleness problem into an availability problem at 9k queries/second, and choosing not to do that is the trade.
Define the cutover numerically. Writes are untouched by the rebuild - they commit to the primary and land in the outbox - so the only coupling is apply throughput. If catch-up applies slower than the 4k events/second arriving, it never converges. The cutover criterion is that lag is measurably decreasing and below a stated threshold, not that the backfill loop reached the end of its range.

Worked solution (30 min)

Write the rebuild sequence: snapshot source, populate the second index, attach the live tail, verify, swap alias, retire the old index - naming what is reversible at each step.
Define the per-partition position record and the single watermark derived from it, and show what each would report when one partition stalls.
Write the idempotent apply rule using the document's recorded aggregate_version, and replay one event twice against it.
State the cutover threshold in lag terms and the measurement that shows convergence rather than completion.

Expected result: A parallel index populated from resource_revision snapshots plus the live tail, cut over by an alias swap that is reversible. Position is stored per log partition and the published watermark comes from the least advanced one. Apply is idempotent on the document's recorded aggregate_version, so backfill and tail can overlap without a freeze. Results carry the watermark and writes return their version so clients can detect staleness. Cutover happens when lag is decreasing and under the stated threshold.

Verify it yourself

Applying the same event twice leaves the document unchanged.
The published watermark never advances past the least advanced partition.
Pointing the alias back restores previous behaviour with no reindex.
Catch-up rate exceeds the 4k events/second arrival rate before cutover, shown by the lag trend rather than by the loop finishing.

Follow-up

The rebuilt index disagrees with the primary tables for 300 documents. Which is authoritative, and how do you decide without freezing writes?
The rebuild doubles load on the log and pushes the projection p99 from 120 ms to 400 ms. What do you throttle, and which signal sets how much?
Clients start polling until the watermark passes their write. What does that do at 9k queries/second, and what do you offer instead?

Design the async export contract a client can resume safely

Hard · async operations · idempotency · leases

A tenant asks for a CSV of every resource. The work runs for minutes on the worker fleet through a job_run row carrying a lease, an attempt count and a dedupe_key, far past the edge's 400 ms budget. Callers are a browser that polls and a script that walks away and checks later. Specify what the submit call returns, the operation resource and its states, how a duplicate submit is handled, how a client learns about completion, what cancellation means given that a lease can expire mid-run, and how the result is fetched and when it expires.

Approach

Split the API in two. Submit returns 202 with an operation id and a location to poll, and never blocks on the work. The operation is a real resource with its own lifecycle - queued, running, succeeded, failed, cancelled - plus attempt, a monotonic progress figure, and a terminal error drawn from the same code taxonomy the synchronous endpoints use, so a client needs one error vocabulary rather than two.
Deduplicate at submit using job_run.dedupe_key, unique over (job_type, dedupe_key) while status is 'queued' or 'running': a repeat submit of the same logical export returns 200 with the existing operation instead of 202 with a new one, and the partial index deliberately permits a legitimate re-run once the first has finished. Pair it with the request's idempotency key so an HTTP-level retry of the submit is exact rather than merely similar.
Tell the poller how to poll: Retry-After on the polling response, a minimum interval enforced at the edge, and a documented maximum lifetime after which an operation is reaped. Polling is the contract of record; the webhook is the fast path, and both must lead to the same terminal state, so a client that receives the completion event and then polls anyway sees no contradiction.
Be exact about cancellation. A cancel request records intent; it cannot stop work already executing. The handler reads the flag at checkpoints, and because a lease expires on a clock that cannot distinguish a dead worker from a slow one, a second copy may start after the cancel was recorded - so the handler re-reads the flag immediately after claiming the lease. 'cancelled' becomes terminal only when no lease is outstanding; reporting it earlier shows a client a stopped job while a worker is still writing output.
Make the handler safe to run twice, because the lease guarantees that it will be. Write output to a deterministic object key derived from the operation id so a second copy overwrites its own work instead of appending a second file, and record completion with a conditional update that only the copy holding the current lease can win.
Treat result fetch as a separate authorised read: a short-lived signed URL, the tenant checked when it is issued rather than only when the file was produced, and a documented retention after which the operation remains terminal but the bytes are gone - a state the client must be able to tell apart from a failure.

Follow-up

An operation has said 'running' for 40 minutes and the worker is gone. What does the client see, and which columns in job_run decide that?
Two tenants each submit 50 exports at once. What in this contract stops one of them delaying the other?
The customer wants the export emailed instead. What changes, and what becomes harder to make exactly-once?

Edge instances grow 400 MB per hour until the nightly restart

Hard · memory leak · heap analysis · metric cardinality · gc

Edge API instances start at 700 MB resident and grow about 400 MB/hour; a nightly rolling restart has hidden it for weeks. Growth continues unchanged when request rate halves overnight, p99 degrades in the last hours before an instance is recycled, and heap used immediately after a forced full GC rises monotonically. The service holds no product state. Name the discriminating measurement that separates the plausible causes, give the most likely cause, and give the fix and how you would verify it.

Approach

Separate resident memory from live heap first, because they fail differently. Resident size can grow from fragmentation, native buffers or thread stacks while the heap is flat; heap used after a full GC rising monotonically is the measurement that says objects are reachable and not being released. You already have it, so this is retention, not fragmentation, and that closes off half the candidate list.
Use the rate's independence from traffic as the discriminator. Growth that continues at half the request rate rules out per-request objects that are merely slow to collect and points at a structure that grows with distinct values observed rather than with call volume. Write the candidates that have that property: a metrics registry keyed on a high-cardinality label, an unevicted cache, an interner, a per-key lock map.
Take two heap snapshots an hour apart and diff by retained size, reading the dominator tree, not by allocation count or instance count. Expect one root holding a map with millions of entries, then follow the reference chain to the code that inserts and never removes. Allocation profilers point at churn, which is the wrong signal here.
The candidate that fits this service is an observability label carrying an identifier, such as a request path recorded before templating so that /v1/resources/48213 becomes its own metric series. That grows with distinct ids seen, is independent of rate, and explains the late p99 degradation, since GC cost rises with the size of the live set.
Fix by bounding cardinality at the source: template the path to /v1/resources/{id} before it becomes a label, move tenant id from a label to a log field or an exemplar, and cap the registry with a bounded map that evicts. Add a cardinality ceiling that fails loudly in a lower environment rather than growing quietly in production.
Verify with a soak rather than a restart. Hold one instance out of the nightly recycle for 48 hours with the fix and compare post-GC heap and series count against an unfixed control taking the same traffic.

Follow-up

Post-GC heap is now flat but resident size still creeps. What are you looking at, and does it matter?
How would you have detected this before an OOM, given the nightly restart masked the trend?
That label is what makes one dashboard useful. How do you keep the dashboard and lose the leak?

Suggested practice checklist

Roughly ninety minutes on weeknights with one longer weekend block. The plan cuts scope rather than compressing everything, on the assumption that one thing finished per night beats four half-started.

Day 1 — Fix the scope and take a cold baseline

Read the role description and write the three things the loop will almost certainly test, then write an explicit not-doing list and keep it visible all week.
Take one twenty-five-minute coding problem and one fifteen-minute design prompt cold, and write the single sentence naming what blocked each, because those two sentences decide where the remaining evenings go.
Set the week's rule: one thing finished every night, including the night you only have forty minutes.

Deliverable: A one-page scope with a not-doing list and two cold attempts, each carrying one sentence on what blocked it.

Day 2 — One pattern, written three times from blank

Choose the single pattern most likely to appear in your loop and write it three times from an empty file rather than editing the previous attempt.
On the third pass, write the invariant as a comment before the loop body and the complexity before the first line of code.
Stop at ninety minutes even if the third version is imperfect, and write the one thing you would fix given another hour.

Deliverable: Three independent implementations of the same pattern plus a note on what changed between them.

Day 3 — One design, only to the depth you can defend

Take one system shape and go only as far as requirements, interface and data model, refusing to draw a box you could not survive a follow-up about.
Attach one number to each non-functional requirement, deriving it rather than asserting it, and write the assumption the number rests on.
Write the one tradeoff you are choosing against and the observation that would make you reverse it.

Deliverable: One design at interface-and-schema depth with derived numbers and one written reversible tradeoff.

Day 4 — Only the fundamentals you will have to defend

Write, in under two hundred words each, the answers to the two questions that follow almost any implementation: why this structure and not the obvious alternative, and what happens to this code at a hundred times the input.
Write what an index actually costs: faster lookups on the indexed columns against a write that now maintains a second structure, plus the cases where the planner declines to use it anyway, low selectivity, or a predicate wrapping the column in a function.
Delete any answer you cannot deliver aloud in under a minute, since an answer that needs reading is not an answer you have.

Deliverable: Three written answers, each under two hundred words and each timed aloud.

Day 5 — Your own work, timed

Write a ninety-second and a four-minute version of your main project and time both aloud rather than reading them.
Prepare the two follow-ups that always come: what you would do differently, and how you knew it worked.
Put one number in the first sentence and be ready to say exactly where it came from and what it excludes.

Deliverable: Two timed narratives with one defensible number in the opening line.

Day 6 — The one full rehearsal, in the weekend block

Run a sixty-minute mock covering a coding round and a design round in one sitting with no break, because sustained attention is the thing evenings have not trained.
Immediately afterwards, and before hearing any feedback, write the three moments you lost the thread.
Spend the rest of the block only on those three moments, and on nothing you merely feel shaky about.

Deliverable: Mock notes naming three failure moments with a specific fix written under each.

Day 7 — Taper

Write the twenty-minute warm-up you will actually do on the morning: one problem you can already solve from a blank file, one design you can narrate, and nothing you have never seen.
Re-read only your own notes from this week and open no new material.
Write the logistics down: the editor or shared document you will be working in, whether execution and lookups are permitted, and the sentence you will use when you do not know something.

Deliverable: A one-page card holding the design structure, the project numbers, and the logistics.

Outage answers are graded on instruments

Nobody is scoring your stamina at three in the morning. What carries weight is which signal told you something was wrong, what you measured before touching anything, what you rolled back versus what you fixed forward, and why you picked one. 'We restarted it and it went away' is a story about not knowing.

What did you learn from your role at your previous position or internship?

Medium · behavioural and engineering judgement

Approach

Give the blast radius: what could have broken, and what you measured.
Pick a story where you made the decision, not one where you watched it.
Close with what you would do differently, concretely.

Follow-up

What did you decide not to do, and why?
What would you do differently if you ran that again?

How do you approach your work and manage your daily responsibilities?

Medium · behavioural and engineering judgement

Approach

Name the disagreement and how you resolved it with evidence.
Give the blast radius: what could have broken, and what you measured.
State the situation in two sentences and spend the rest on the reasoning.

Follow-up

What would you do differently if you ran that again?
How did you know your change caused the improvement?

Ship under a deadline and bound the debt you chose

Medium · pagination · technical debt · tradeoffs

You have four days to ship a tenant-facing listing endpoint. The version you would defend uses keyset pagination over (tenant_id, status, updated_at DESC, resource_id DESC); the version you can finish uses LIMIT/OFFSET with no matching index. Describe a deadline call you actually made of this shape: what you shipped, what you knowingly deferred, how you bounded the damage with a mechanism rather than an intention, and the specific numeric condition that would force the follow-up. Name who you told and where you wrote it down.

Approach

Name the deferred failure precisely instead of calling it slow. OFFSET n makes the database produce and discard n rows, so cost grows with page depth; without an index matching the sort, every matching row is read and sorted before the limit applies; and rows inserted between two page fetches shift across the boundary so items are skipped or repeated with nothing in the response to signal it.
Bound the blast radius with something mechanical rather than a promise: cap maximum page depth, cap page size, restrict the endpoint to one internal caller, or keep it behind a flag. State which failure each cap removes and which it leaves standing.
Attach a number to the trigger and wire it to an alarm: the first tenant crossing N resources, or the endpoint's p99 crossing its share of the 400 ms budget, so the debt announces itself instead of waiting to be remembered.
Write it where the next engineer looks, which is the code and the ticket, not a chat message: what was deferred, why, the cap, and the trigger.
Report what actually happened in your real example, including the case where the trigger never fired and the debt was correctly never repaid.

Follow-up

At what page depth does the offset version breach your latency budget, given your page size and row counts?
What breaks first when you switch to keyset pagination later, and what does a client holding an old page token see?
Who would have overruled you if you had asked for two more days, and did you ask?